LeetCode #2045 — HARD

Second Minimum Time to Reach Destination

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

A city is represented as a bi-directional connected graph with n vertices where each vertex is labeled from 1 to n (inclusive). The edges in the graph are represented as a 2D integer array edges, where each edges[i] = [u_i, v_i] denotes a bi-directional edge between vertex u_i and vertex v_i. Every vertex pair is connected by at most one edge, and no vertex has an edge to itself. The time taken to traverse any edge is time minutes.

Each vertex has a traffic signal which changes its color from green to red and vice versa every change minutes. All signals change at the same time. You can enter a vertex at any time, but can leave a vertex only when the signal is green. You cannot wait at a vertex if the signal is green.

The second minimum value is defined as the smallest value strictly larger than the minimum value.

For example the second minimum value of [2, 3, 4] is 3, and the second minimum value of [2, 2, 4] is 4.

Given n, edges, time, and change, return the second minimum time it will take to go from vertex 1 to vertex n.

Notes:

You can go through any vertex any number of times, including 1 and n.
You can assume that when the journey starts, all signals have just turned green.

Example 1:

Input: n = 5, edges = [[1,2],[1,3],[1,4],[3,4],[4,5]], time = 3, change = 5
Output: 13
Explanation:
The figure on the left shows the given graph.
The blue path in the figure on the right is the minimum time path.
The time taken is:
- Start at 1, time elapsed=0
- 1 -> 4: 3 minutes, time elapsed=3
- 4 -> 5: 3 minutes, time elapsed=6
Hence the minimum time needed is 6 minutes.

The red path shows the path to get the second minimum time.
- Start at 1, time elapsed=0
- 1 -> 3: 3 minutes, time elapsed=3
- 3 -> 4: 3 minutes, time elapsed=6
- Wait at 4 for 4 minutes, time elapsed=10
- 4 -> 5: 3 minutes, time elapsed=13
Hence the second minimum time is 13 minutes.

Example 2:

Input: n = 2, edges = [[1,2]], time = 3, change = 2
Output: 11
Explanation:
The minimum time path is 1 -> 2 with time = 3 minutes.
The second minimum time path is 1 -> 2 -> 1 -> 2 with time = 11 minutes.

Constraints:

2 <= n <= 10⁴
n - 1 <= edges.length <= min(2 * 10⁴, n * (n - 1) / 2)
edges[i].length == 2
1 <= u_i, v_i <= n
u_i != v_i
There are no duplicate edges.
Each vertex can be reached directly or indirectly from every other vertex.
1 <= time, change <= 10³

Step 01

Brute Force Baseline

Problem summary: A city is represented as a bi-directional connected graph with n vertices where each vertex is labeled from 1 to n (inclusive). The edges in the graph are represented as a 2D integer array edges, where each edges[i] = [ui, vi] denotes a bi-directional edge between vertex ui and vertex vi. Every vertex pair is connected by at most one edge, and no vertex has an edge to itself. The time taken to traverse any edge is time minutes. Each vertex has a traffic signal which changes its color from green to red and vice versa every change minutes. All signals change at the same time. You can enter a vertex at any time, but can leave a vertex only when the signal is green. You cannot wait at a vertex if the signal is green. The second minimum value is defined as the smallest value strictly larger than the minimum value. For example the second minimum value of [2, 3, 4] is 3, and the second minimum

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

5
[[1,2],[1,3],[1,4],[3,4],[4,5]]
3
5

Example 2

2
[[1,2]]
3
2

Core Insight

What unlocks the optimal approach

How much is change actually necessary while calculating the required path?
How many extra edges do we need to add to the shortest path?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2045: Second Minimum Time to Reach Destination
class Solution {
    public int secondMinimum(int n, int[][] edges, int time, int change) {
        List<Integer>[] g = new List[n + 1];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (int[] e : edges) {
            int u = e[0], v = e[1];
            g[u].add(v);
            g[v].add(u);
        }
        Deque<int[]> q = new LinkedList<>();
        q.offerLast(new int[] {1, 0});
        int[][] dist = new int[n + 1][2];
        for (int i = 0; i < n + 1; ++i) {
            Arrays.fill(dist[i], Integer.MAX_VALUE);
        }
        dist[1][1] = 0;
        while (!q.isEmpty()) {
            int[] e = q.pollFirst();
            int u = e[0], d = e[1];
            for (int v : g[u]) {
                if (d + 1 < dist[v][0]) {
                    dist[v][0] = d + 1;
                    q.offerLast(new int[] {v, d + 1});
                } else if (dist[v][0] < d + 1 && d + 1 < dist[v][1]) {
                    dist[v][1] = d + 1;
                    if (v == n) {
                        break;
                    }
                    q.offerLast(new int[] {v, d + 1});
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < dist[n][1]; ++i) {
            ans += time;
            if (i < dist[n][1] - 1 && (ans / change) % 2 == 1) {
                ans = (ans + change) / change * change;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2045: Second Minimum Time to Reach Destination
// Auto-generated Go example from java.
func exampleSolution() {
}
// Reference (java):
// // Accepted solution for LeetCode #2045: Second Minimum Time to Reach Destination
// class Solution {
//     public int secondMinimum(int n, int[][] edges, int time, int change) {
//         List<Integer>[] g = new List[n + 1];
//         Arrays.setAll(g, k -> new ArrayList<>());
//         for (int[] e : edges) {
//             int u = e[0], v = e[1];
//             g[u].add(v);
//             g[v].add(u);
//         }
//         Deque<int[]> q = new LinkedList<>();
//         q.offerLast(new int[] {1, 0});
//         int[][] dist = new int[n + 1][2];
//         for (int i = 0; i < n + 1; ++i) {
//             Arrays.fill(dist[i], Integer.MAX_VALUE);
//         }
//         dist[1][1] = 0;
//         while (!q.isEmpty()) {
//             int[] e = q.pollFirst();
//             int u = e[0], d = e[1];
//             for (int v : g[u]) {
//                 if (d + 1 < dist[v][0]) {
//                     dist[v][0] = d + 1;
//                     q.offerLast(new int[] {v, d + 1});
//                 } else if (dist[v][0] < d + 1 && d + 1 < dist[v][1]) {
//                     dist[v][1] = d + 1;
//                     if (v == n) {
//                         break;
//                     }
//                     q.offerLast(new int[] {v, d + 1});
//                 }
//             }
//         }
//         int ans = 0;
//         for (int i = 0; i < dist[n][1]; ++i) {
//             ans += time;
//             if (i < dist[n][1] - 1 && (ans / change) % 2 == 1) {
//                 ans = (ans + change) / change * change;
//             }
//         }
//         return ans;
//     }
// }

# Accepted solution for LeetCode #2045: Second Minimum Time to Reach Destination
class Solution:
    def secondMinimum(
        self, n: int, edges: List[List[int]], time: int, change: int
    ) -> int:
        g = defaultdict(set)
        for u, v in edges:
            g[u].add(v)
            g[v].add(u)
        q = deque([(1, 0)])
        dist = [[inf] * 2 for _ in range(n + 1)]
        dist[1][1] = 0
        while q:
            u, d = q.popleft()
            for v in g[u]:
                if d + 1 < dist[v][0]:
                    dist[v][0] = d + 1
                    q.append((v, d + 1))
                elif dist[v][0] < d + 1 < dist[v][1]:
                    dist[v][1] = d + 1
                    if v == n:
                        break
                    q.append((v, d + 1))
        ans = 0
        for i in range(dist[n][1]):
            ans += time
            if i < dist[n][1] - 1 and (ans // change) % 2 == 1:
                ans = (ans + change) // change * change
        return ans

// Accepted solution for LeetCode #2045: Second Minimum Time to Reach Destination
/**
 * [2045] Second Minimum Time to Reach Destination
 *
 * A city is represented as a bi-directional connected graph with n vertices where each vertex is labeled from 1 to n (inclusive). The edges in the graph are represented as a 2D integer array edges, where each edges[i] = [ui, vi] denotes a bi-directional edge between vertex ui and vertex vi. Every vertex pair is connected by at most one edge, and no vertex has an edge to itself. The time taken to traverse any edge is time minutes.
 * Each vertex has a traffic signal which changes its color from green to red and vice versa every change minutes. All signals change at the same time. You can enter a vertex at any time, but can leave a vertex only when the signal is green. You cannot wait at a vertex if the signal is green.
 * The second minimum value is defined as the smallest value strictly larger than the minimum value.
 *
 * 	For example the second minimum value of [2, 3, 4] is 3, and the second minimum value of [2, 2, 4] is 4.
 *
 * Given n, edges, time, and change, return the second minimum time it will take to go from vertex 1 to vertex n.
 * Notes:
 *
 * 	You can go through any vertex any number of times, including 1 and n.
 * 	You can assume that when the journey starts, all signals have just turned green.
 *
 *  
 * Example 1:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/09/29/e1.png" style="width: 200px; height: 250px;" /> &emsp; &emsp; &emsp; &emsp;<img alt="" src="https://assets.leetcode.com/uploads/2021/09/29/e2.png" style="width: 200px; height: 250px;" />
 * Input: n = 5, edges = [[1,2],[1,3],[1,4],[3,4],[4,5]], time = 3, change = 5
 * Output: 13
 * Explanation:
 * The figure on the left shows the given graph.
 * The blue path in the figure on the right is the minimum time path.
 * The time taken is:
 * - Start at 1, time elapsed=0
 * - 1 -> 4: 3 minutes, time elapsed=3
 * - 4 -> 5: 3 minutes, time elapsed=6
 * Hence the minimum time needed is 6 minutes.
 * The red path shows the path to get the second minimum time.
 * - Start at 1, time elapsed=0
 * - 1 -> 3: 3 minutes, time elapsed=3
 * - 3 -> 4: 3 minutes, time elapsed=6
 * - Wait at 4 for 4 minutes, time elapsed=10
 * - 4 -> 5: 3 minutes, time elapsed=13
 * Hence the second minimum time is 13 minutes.      
 *
 * Example 2:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/09/29/eg2.png" style="width: 225px; height: 50px;" />
 * Input: n = 2, edges = [[1,2]], time = 3, change = 2
 * Output: 11
 * Explanation:
 * The minimum time path is 1 -> 2 with time = 3 minutes.
 * The second minimum time path is 1 -> 2 -> 1 -> 2 with time = 11 minutes.
 *  
 * Constraints:
 *
 * 	2 <= n <= 10^4
 * 	n - 1 <= edges.length <= min(2 * 10^4, n * (n - 1) / 2)
 * 	edges[i].length == 2
 * 	1 <= ui, vi <= n
 * 	ui != vi
 * 	There are no duplicate edges.
 * 	Each vertex can be reached directly or indirectly from every other vertex.
 * 	1 <= time, change <= 10^3
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/second-minimum-time-to-reach-destination/
// discuss: https://leetcode.com/problems/second-minimum-time-to-reach-destination/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn second_minimum(n: i32, edges: Vec<Vec<i32>>, time: i32, change: i32) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2045_example_1() {
        let n = 5;
        let edges = vec![vec![1, 2], vec![1, 3], vec![1, 4], vec![3, 4], vec![4, 5]];
        let time = 3;
        let change = 5;

        let result = 13;

        assert_eq!(Solution::second_minimum(n, edges, time, change), result);
    }

    #[test]
    #[ignore]
    fn test_2045_example_2() {
        let n = 2;
        let edges = vec![vec![1, 2]];
        let time = 3;
        let change = 2;

        let result = 11;

        assert_eq!(Solution::second_minimum(n, edges, time, change), result);
    }
}

// Accepted solution for LeetCode #2045: Second Minimum Time to Reach Destination
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2045: Second Minimum Time to Reach Destination
// class Solution {
//     public int secondMinimum(int n, int[][] edges, int time, int change) {
//         List<Integer>[] g = new List[n + 1];
//         Arrays.setAll(g, k -> new ArrayList<>());
//         for (int[] e : edges) {
//             int u = e[0], v = e[1];
//             g[u].add(v);
//             g[v].add(u);
//         }
//         Deque<int[]> q = new LinkedList<>();
//         q.offerLast(new int[] {1, 0});
//         int[][] dist = new int[n + 1][2];
//         for (int i = 0; i < n + 1; ++i) {
//             Arrays.fill(dist[i], Integer.MAX_VALUE);
//         }
//         dist[1][1] = 0;
//         while (!q.isEmpty()) {
//             int[] e = q.pollFirst();
//             int u = e[0], d = e[1];
//             for (int v : g[u]) {
//                 if (d + 1 < dist[v][0]) {
//                     dist[v][0] = d + 1;
//                     q.offerLast(new int[] {v, d + 1});
//                 } else if (dist[v][0] < d + 1 && d + 1 < dist[v][1]) {
//                     dist[v][1] = d + 1;
//                     if (v == n) {
//                         break;
//                     }
//                     q.offerLast(new int[] {v, d + 1});
//                 }
//             }
//         }
//         int ans = 0;
//         for (int i = 0; i < dist[n][1]; ++i) {
//             ans += time;
//             if (i < dist[n][1] - 1 && (ans / change) % 2 == 1) {
//                 ans = (ans + change) / change * change;
//             }
//         }
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.